Question

Cho A = 2 + 2² + 2³ ...+ 2²⁰ . Chứng minh rằng:
a) A chia hết cho 2
b) A chia hết cho 3
c) A chia hết cho 5

Em đang rất gấp ạ

 

Answer 1

a: Ta có: \(A=2+2^2+2^3+...+2^{20}\)

\(=2\left(1+2+2^2+...+2^{19}\right)⋮2\)

b: Ta có: \(A=2+2^2+2^3+...+2^{20}\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{19}\left(1+2\right)\)

\(=3\cdot\left(2+2^3+...+2^{19}\right)⋮3\)

Answer 2

c) tham khảo:

M = 2 + 22 + 23 + ... + 220
= ( 2 + 22 + 23 + 24 ) + ( 25 + 26 + 27 + 28 ) + ... + ( 217 + 218 + 219 + 220 )
= 2 . ( 1 + 2 + 22 + 23 ) + 25 . ( 1 + 2 + 22 + 23 ) + ... + 217 . ( 1 + 2 + 22 + 23 )
= 2 . 15 + 25 . 15 + ... + 217 .15
= 15 . 2 ( 1 + 24 + ... + 216 )
= 3 . 5 . 2 ( 1 + 24 + ... + 216 ) \(⋮\) 5

Answer 3

Lời giải:
a. 

$A=2(1+2^1+2^2+...+2^{19})\vdots 2$

b. 

$A=(2+2^2)+(2^3+2^4)+.....+(2^{19}+2^{20})$

$=2(1+2)+2^3(1+2)+....+2^{19}(1+2)$

$=2.3+2^3.3+...+2^{19}.3$

$=3(2+2^3+...+2^{19})\vdots 3$

c.

$A=(2+2^2+2^3+2^4)+(2^5+2^6+2^7+2^8)+....+(2^{17}+2^{18}+2^{19}+2^{20})$

$=2(1+2+2^2+2^3)+2^5(1+2+2^2+2^3)+....+2^{17}(1+2+2^2+2^3)$

$=2.15+2^5.15+....2^{17}.15$
$=15(2+2^5+...+2^{17})$
$=5.3.(2+2^5+...+2^{17})\vdots 5$

Answer 4

a) Ta có: 2, 2², 2³,...2²⁰ đều chia hết cho 2

=> A=\(2^2+2^3+...+2^{20}⋮2\)

b) \(A=2+2^2+2^3+...+2^{20}=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{19}\left(1+2\right)=3.2+3.2^3+...+3.2^{19}=3\left(2+2^3+...+2^{19}\right)⋮3\)c) \(A=2+2^2+2^3+...+2^{20}=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+...2^{17}\left(1+2+2^2+2^3\right)=15\left(2+2^5+...+2^{17}\right)⋮15\Rightarrow⋮5\Rightarrow A⋮5\)